# Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$

I got this problem:

Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$.
(Hint: the solution involves limits at infinity)

I tried to prove it (by contradiction) but I failed. Thanks on any help.

DISCLAIMER: This answer is incorrect as it stands, namely in the $$k>0$$ section is claims that: $$f(0)=f(x-x)\geq f(x)-x f(f(x))$$ implies $$f(x), which is false.

This disclaimer will stay here until the problem is fixed.

From the condition $$f(0)>0$$ it follows that with $$x=0$$ we have $$f(0+y)-y f(f(0))\geq f(0)>0$$ so it follows that $$f(y)>ky$$ where $$k=f(f(0))$$ is some non-zero constant. It has to be non-zero since otherwise $$0=k=f(f(x))>k f(x)=0$$. Now we have to divide into cases.

## If $$k>0$$

If $$k>0$$ it follows that $$x f(f(x))>kx f(x)>k^2 x^2\geq 0$$. So in that case we have $$\underbrace{f(x-x)}_{f(0)}\geq f(x)-\underbrace{x f(f(x))}_{\text{strictly greater than }0}\\ \iff\\ f(x) But the above statement is absurd by all means. For one thing $$f(0) and moreover $$f(0)>f(x)>kx\rightarrow\infty$$ which is also absurd.

## If $$k<0$$

If we apply the beginning of what Denis showed, namely that $$f(x)\rightarrow\infty$$ for $$x\rightarrow\infty$$, then we get in addition to that from my analysis that for $$k<0$$ we have $$f(-x)\geq -kx\rightarrow\infty$$ for $$x\rightarrow\infty$$. So we see that $$f(0)=f(-x+x)\geq f(-x)+x f(f(-x))\rightarrow\infty$$ for $$x\rightarrow\infty$$ too. A contradiction since we have just shown that the constant $$f(0)$$ is greater than or equal to an expression that tends to infinity. So $$k<0$$ is impossible too.

• If you put x = 0, then the LHS in the first inequality you gave should be f(0) - yf(f(0)) – TenaliRaman Sep 24 '14 at 8:43
• @TenaliRaman: No! The LHS is still $f(0+y)-y f(f(0))$ even if $y=0$, and the statement $f(y)>ky$ holds for ANY $y$ including $y=0$. – String Sep 24 '14 at 8:49
• @TenaliRaman: But I spotted another serious problem in my analysis which I just pointed out in my answer above ... – String Sep 24 '14 at 8:53
• @String: Maybe we can suppose that $\forall x\in\Bbb{R}, f(f(x))\leq 0$ and from that reach to a contradiction and get that $\exists z\in\Bbb{R}$ such that $f(f(z))>0$ which will be something like the $k$ that you mentioned. – MathNerd Sep 24 '14 at 9:04
• @Saita: I finished my argument! – String Sep 24 '14 at 10:32

Let $$a=f(0)>0$$, $$b=f(a)$$ and $$c=f(b)$$.

We know that $$f(0+b)\geq a+b^2$$, so we get $$c>0$$.

We also have for all $$x$$, $$f(x+a)\geq f(a)+cx$$, so in particular $$\lim_{x\to\infty} f(x)=\infty$$ and more precisely $$f(x)=\Omega(x)$$ when $$x\to\infty$$.

This means that for $$x$$ big enough, $$f(x)>0$$, and therefore, $$f(x+1)\geq f(x)+f(f(x))>f(f(x))$$. This gets us $$f(f(x-1)).

Moreover, for $$x$$ big enough, we have $$f(x-1)>x$$ (since $$f(x+1)=\Omega(x^2)$$).

Let us choose $$x_0$$ big enough, and define $$x_{n+1}=f(x_n-1)$$.

The sequence $$(x_n)$$ is strictly increasing, and for all $$n$$, we have $$f(x_{n+1}). We reached a a contradiction with $$\lim_{x\to\infty} f(x)=\infty$$.

• So to sum up you have shown $f(x+a)\geq cx+b$ where $c>0$. Very nice! How do you conclude from that? – String Sep 24 '14 at 9:29
• I found the end :) – Denis Sep 24 '14 at 10:11
• It took me a while to absorb, but it makes sense now. Perhaps you should state the inequalities as $f(x)>f(f(x-1))$ and $f(x-1)>x$ respectively to make the reading more direct, but that is just an opinion. The first part of your answer showing that $f(x)\rightarrow\infty$ provided the last piece of the puzzle for my version of it BTW :) – String Sep 24 '14 at 10:40
• I smoothed it accordingly. – Denis Sep 24 '14 at 10:47

By what Denis said we get that $lim_{x\to\infty}f(x)=\infty$ And so $lim_{x\to\infty}f(f(x))=\infty$.

Now we will reach a contradiction by using the definition of limits at infinity:

Because $lim_{x\to\infty}f(x)=\infty$ and $lim_{x\to\infty}f(f(x))=\infty$. We get that:
$\forall 0<M_1,\exists 0<N_1$ such that $\forall N_1<x, M_1<f(x)$ and that
$\forall 0<M_2,\exists 0<N_2$ such that $\forall N_2<x, M_2<f(f(x))$

Now take $M_1=M_2=1$ and we will get that $\exists 0<N_1$ such that $\forall N_1<x, 1<f(x)$ and that $\exists 0<N_2$ such that $\forall N_2<x, 1<f(f(x))$

Take $x=max (N_1+1,N_2+1)$, then $N_1,N _2<x$ and so we get that $1<f(x)$ and that $1<f(f(x))$

Now take $y=max(\frac{x+1}{f( f( x))-1},N_2-x)$ which implies that $N_2< x+y+1\leq y f(f(x ))$ and so $1<f(f(x+y+1))$, Now $f (x+y)\geq f(x )+yf(f(x))\geq x+y+1$ And hence

$f(f(x+y))\geq f(x+y+1)+(f(x+y)-(x+y+1))f(f(x+y+1)) \geq f(x +y+1)\geq f(x+y)+f(f(x+y))\geq f(x)+yf(f(x))+f(f(x+y))>f(f(x+y))$

And so we get that $0>0$ which is a contradiction.